The incorrect way: just superpose the solutions, compute Poynting flux
Far from the dipole we have the leading dipole solution $$\vec{A}_d = -\frac{\mu_0 \omega}{4 \pi r} e^{i \omega(t-r)} \vec{d} +c.c., \phi_d = - \frac{\mu_0 \omega}{4\pi r} e^{i \omega(t-r)}\vec{d}\cdot\hat{r}+c.c.$$ Here I am using $c=1$ units, $c.c.$ stands for complex conjugate, $\hat{r} = (x,y,z)/r$ is the unit distance vector, and the formal complex dipole vector reads $\vec{d} = 2Lq(1,i,0)$. The wave potential is then most conveniently expressed in the Gibbs gauge $\phi=0$: $$\vec{A}_W = \frac{\vec{a}}{\omega} e^{i\omega(t-z)}+c.c., \phi_W = 0$$ Here $\vec{a} = E_{ext} (1,i,0)$.
By adding these potentials, it is then easy to obtain the total $\vec{E}$ and $\vec{B}$. The Poynting flux vector can then be computed simply by "turning the crank". I am only showing the final result for the radial flux through the angle element $d \theta$, integrated over $\varphi$ ($r,\theta,\varphi$ standard polar coordinates)
$$\int_0^{2\pi}\frac{r^2 \sin\theta}{\mu_0} \hat{r}\cdot \left( \vec{E}\times\vec{B}\right) d\varphi = F_{dip.}+F_{cross.}+F_{wave}$$ $F_{dip.}$ and $F_{wav.}$ are the same terms as with the dipole on its own and the wave on its own. The $F_{cross}$ term is new and reads: $$F_{cross} = \frac{1}{2} E_{ext.}L q r \omega ^2 \sin \theta \left(\cos \frac{\theta }{2}-\sin \frac{\theta }{2}\right)^4 \cos (r \omega (\sin \theta -1))$$ It has a weird behavior that I find hard to understand. The total flux $\int F_{cross} d\theta$ changes sign and diverges as $r\to\infty$. There is no meaningful way to "average out" this behavior.
So what is the meaning of this? After some thought, I believe that $F_{cross}$ balances out the energy needed to keep the system in a steady state for an indefinite time, which follows from implicit assumptions. The rotating dipole is not really moving in the external field, no equations of motion are being solved - so one cannot get consistent momentum-energy balances.
We could then instead choose to solve equations of motion. This would require allowing for a dynamical and independent $\omega(t)$ of the dipole (or rather phase $\varphi(t)$). The evolution would depend (nonlinearly) on the masses of the charges, and the solution would be nonstationary. As a result, one should get a self-consistent radiative field and also the correct balance reflecting $P_{ext.}$ in the fluxes. I think that verifying this would amount to a neat Bachellor's or even Master's thesis.
The more fundamental conclusion that I am seeing is that there is little similarity to quantum spontaneous emission in this setup. The external field initially acts coherently on the dipole by matching its frequency, and the emerging radiation will have the same frequency (for a short time), but that is about where the analogy can be extended. There is no net momentum transfer, in particular no $z$-component - so the emerging wave cannot carry any net momentum and has to be isotropic. In other words, it has to be very similar to a classical dipole field.